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 October 17th, 2016, 07:18 PM #1 Senior Member     Joined: Nov 2010 From: Indonesia Posts: 2,000 Thanks: 132 Math Focus: Trigonometry [ASK] Probabilities There are two boxes. Box I contains 5 red balls and 4 white balls. Box II contains 3 red balls and 6 white balls. A ball is taken randomly from box I and put into box II. Then, from box II, two balls are taken randomly. What is the probability of both balls taken from box II are red? I only know that the probability of getting red ball from box I is $\displaystyle \frac{1}{9}$. How to do the rest?
 October 17th, 2016, 07:53 PM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 2,100 Thanks: 1093 The initial state is $(5,4),~(3,6)$ If a red ball is chosen and transferred the state becomes $(4,4),~(4,6)$ There is a probability of $\dfrac 5 9$ that this occurs On the other hand if a white ball is chosen and transferred the state becomes $(5,3),~(3,7)$ There is a probability of $\dfrac 4 9$ that this occurs. From $(4,4),~(4,6)$ the probability of choosing 2 red balls from box 2 is $\dfrac{4}{10} \cdot \dfrac{3}{9}$ From $(5,3),~(3,7)$ the probability of choosing 2 red balls from box 2 is $\dfrac{3}{10}\cdot \dfrac{2}{9}$ So the total probability of selecting 2 balls from box 2 is $\dfrac 5 9\cdot \dfrac{4}{10}\cdot \dfrac{3}{9} + \dfrac 4 9 \cdot \dfrac{3}{10}\cdot \dfrac{2}{9} =\dfrac{14}{135}$
 October 17th, 2016, 08:00 PM #3 Senior Member     Joined: Nov 2010 From: Indonesia Posts: 2,000 Thanks: 132 Math Focus: Trigonometry Ugh... It's kinda hard to keep up with the explanation but I think I get it.

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